Averages and range · 平均数与极差
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| mean/miːn/ | 平均数 | píng jūn shù |
| median/ˈmiːdiːən/ | 中位数 | zhōng wèi shù |
| mode/məʊd/ | 众数 | zhòng shù |
| range/reɪndʒ/ | 值域 | zhí yù |
| outlier/ˈaʊtlaɪə/ | 异常值 | yì cháng zhí |
What's "normal"?
- The average salary in a company is 50 000 — but most people earn far less, because one CEO earns millions.
- Which average you choose changes the story. Mean 平均数, median 中位数, and mode 众数 each tell a different truth.
Average choice lab · 平均数选择实验
Choose the average that fits the data situation. · 根据数据情况选择合适的平均数。
The three averages
- Mean $= \dfrac{\text{sum of values}}{\text{how many}}$.
- Median $=$ the middle value when the data is ordered.
- Mode $=$ the most common value.
$4, 7, 7, 2, 5$: ordered $2, 4, 5, 7, 7$. Mean $= \dfrac{25}{5} = 5$; median $= 5$; mode $= 7$.

A box-and-whisker plot shows the five-number summary; the box length is the interquartile range 值域
Find the mean of 4, 7, 7, 2, 5. · 求 4, 7, 7, 2, 5 的算术平均数。
(4 + 7 + 7 + 2 + 5)/5 = 25/5 = 5.
Find the mode of 4, 7, 7, 2, 5. · 求 4, 7, 7, 2, 5 的众数。
7 appears most often. · 7 出现的次数最多。
Find the median of 4, 7, 7, 2, 5. · 求 4, 7, 7, 2, 5 的中位数。
Ordered: 2, 4, 5, 7, 7. The middle (3rd) value is 5. · 排序后:2, 4, 5, 7, 7。中间位置(第3个)的数值是 5。
Match each average or measure to how you find it. · 将每个平均数或度量指标与其计算方法相匹配。
Mean, median and mode are three kinds of average; the range measures spread. · 平均数、中位数和众数是三种平均数;极差用于衡量数据的离散程度。
The range
- Range $=$ largest $-$ smallest. It measures spread.
- Range of $4, 7, 7, 2, 5$: $7 - 2 = 5$.
The range is easily distorted. One extreme value (outlier 异常值) makes the range huge, even if most values are close together. The IQR (next lesson) is more robust.

Mean, median, mode and range of a small data set
Find the range of 4, 7, 7, 2, 5. · 求 4, 7, 7, 2, 5 的极差。
Largest − smallest = 7 − 2 = 5. · 最大值 − 最小值 = 7 − 2 = 5。
The range is easily affected by a single extreme value (outlier). · 极差很容易受到单个极端值(离群值)的影响。
The range uses only the largest and smallest values, so one outlier can make it very large. · 极差仅使用最大值和最小值,因此一个离群值可能导致其变得很大。
Mean from a frequency table
- When values come with frequencies:
- Values $1, 2, 3$ with frequencies $4, 5, 1$: $\;\dfrac{1(4) + 2(5) + 3(1)}{10} = \dfrac{17}{10} = 1.7$.
Values 1, 2, 3 occur with frequencies 4, 5, 1. Find the mean (1 dp). · 值1, 2, 3出现的频率分别为4, 5, 1。求平均值(保留1位小数)。
(1·4 + 2·5 + 3·1)/(4+5+1) = 17/10 = 1.7.
Which average to use?
- Mean: uses all the data, but affected by outliers.
- Median: usually less affected by extreme values than the mean — good for skewed data (like salaries).
- Mode: useful for categorical data (most popular colour, most common shoe size).

Ordering data on a number line makes the median obvious — it's the value in the middle.
Even counts and frequency-table medians
- For $2,5,7,10$, the middle pair is 5 and 7, so median $M=(5+7)/2=6$; do not select only one middle value.
- Values 1,2,3 with frequencies 2,3,1 make six observations. Positions 3 and 4 are both 2, so median 2. Frequency counts determine the middle positions, not the number of table rows.
Values 1, 2, 3 have frequencies 2, 3, 1. Find the median. · 数值 1、2、3 对应的频数分别为 2、3、1。求中位数。
Positions 3 and 4 in the six observations are both 2. · 六个观测值中的第 3 位和第 4 位均为 2。
You've got it
- mean $=$ total ÷ count; median $=$ middle in order; mode $=$ most common
- range $=$ largest − smallest
- frequency-table mean $= \dfrac{\sum(\text{value} \times \text{freq})}{\sum \text{freq}}$